Square roots in rings In a field, if $x$ and $y$ are both square roots of $a$ then either $x = y$ or $x = -y$.  This is easy to prove:
$$(x + y)(x - y) = x^2 - y^2 = a - a = 0$$
So, since we are in a field, either $x + y = 0$ or $x - y = 0$.  
Of course, it is possible that $x$ and $-x$ are not distinct.  
I am wondering about more general rings.  I am not assuming that a ring necessarily has an identity.  
So, I am looking for elements $a$ that have square roots $x$ and $y$ with $x \neq y$ and $x \neq -y$.  I will call these "extra" square roots.  
For an integral domain, the proof still works: there are no extra square roots.
If we drop commutativity then the proof does not work and extra square roots are possible.  For example, in the quaternions there are many square roots of -1.  
So, from now on, I will assume commutativity.  
There appears to be a connection between these extra square roots with both zero divisors and idempotents.  
If $x$, $y$ are both square roots of $a$, it still true that $(x + y)(x - y) = 0$ so if $x \neq y$ and $x \neq -y$ then $x + y$ and $x - y$ are zero divisors.  
However, the reverse is not true, there may be zero divisors but no extra square roots.  This occurs in $\mathbb{Z}_6$.  
Suppose that we have a non-trivial idempotent $e$ and $2$ is cancellable in the ring then $2e - 1$ is a square root of $1$ but $e \neq 1$ and $e \neq -1$.  
However, the reverse is not true $\mathbb{Z}_{15}$ has extra square roots but no idempotents.  That was an error, see comments. 
Thanks to rschwieb, here is a corrected example: $\mathbb{Z}[x]/(x^2−1)$. This has at least three roots of $1$ $( \{1,−1,x\} )$ but only trivial idempotents.
So when $2$ is cancellable, idempotents $\implies$ extra square roots $\implies$ zero divisors but neither converse is true.  
Should I continue?  Is there a hope of finding simple criteria for the presence of extra square roots in commutative rings?
 A: Not sure if it helps but it's too long for a comment.
We have the following results :

Let $p$ be an odd prime, $r\in \mathbb{N}$, $a \in \mathbb{Z}$ s.t. $p$ doesnt divide $a$.
  The equation $x^2 =a$ has $1+(\frac{a}{p})$ solutions in $\mathbb{Z}/p^r\mathbb{Z}$

And :

In $\mathbb{Z}/2^r \mathbb{Z}$, the equation $x^2=a$ (where $a \in \mathbb{Z}$ is odd) has :
  
  
*
  
*if $r=1$, one solution 
  
*if $r=2$, two solutions if $a\equiv 1 \pmod 4$, and zero otherwise 
  
*else, four solutions if $a\equiv 1 \pmod 8$, zero otherwise. 
  

Combining these two results and the chinese remainder theorem, you have a theorem that gives you the number of square roots of a square in $\mathbb{Z}/n\mathbb{Z}$ ; this doesnt really answer your question but it might help.
A: I thought I would complete the analysis of $\mathbb Z_n$ that we had started in the comments, since an assessment of commutative rings in general might be out of reach right now. It turns out the "hard case" for this classification is still elementary.
Case: $n$ not squarefree
Then $\mathbb Z_n$ has nilpotent elements, and therefore $0$ has extra square roots.  
Case: $n$ squarefree and odd
Then it has nontrivial idempotents that will allow your observation to extract extra square roots of $1$.
Case: $n=2m$, $m$ an odd squarefree composite
Then $\mathbb Z_n\cong \mathbb Z_2\oplus \mathbb Z_m$, and the right half will provide nontrivial idempotents which make extra roots of $1$.
Case: $n=2p$, $p$ an odd prime
It turns out that a straightforward computation is sufficient to crack the last case:
Suppose that $2p|x^2-a$ and $2p|y^2-a$. Clearly $2p|(x-y)(x+y)$. Suppose that $2p$ does not divide either $x-y$ or $x+y$. There are two cases: either $2$ divides $x-y$ and $p$ divides $x+y$, or $p$ divides $x-y$ and $2$ divides $x+y$. The analysis of both is the same, so we'll do the former one.
Then $2r=x-y$ and $ps=x+y$, and adding we get $2r+ps=2x$. But this says that $2|ps$, and since $p$ is an odd prime, $2|s$. But now $2p|x+y$ since $2p|ps=x+y$. In this case, then, $x\equiv -y\pmod n$.
Following the same logic in the latter case, $x\equiv y\pmod n$.
This shows that elements don't have extra square roots in $\mathbb Z_{2p}$ for odd primes $p$, and that completes the picture for all $n>1$.
There is probably a fancier way of pointing this out, but this is the first one that occurred to me.
A: Commutative rings can get really ugly and hard to study if you consider all of them at once. Even for finite ones, there might be strange things to happen. Thus, I would first try to figure out nice classes, e.g. $\mathbb{Z}/n\mathbb{Z}$; can you relate the existence of extra roots to $n$ being square free or not for example? Then form there, you might want to extend to Galois rings or other known classes of rings.
Furthermore, you might want to reduce the problem. Do you really want to ask if any element has more than two roots, or would it be enough to look at roots of $1$ first? Note that every unit that has more than two roots will yield a root of $1$ which is not $\pm 1$.
Another nice thing to look at are rings or characteristic two (maybe $\mathbb{F}[x]/\langle p \rangle$ for some polynomial $p$ and $\mathbb{F}$ a field of characteristic two). Here, you have
$$x^2 - y^2 = (x-y)^2,$$
so every element with more than one square root will give a non-trivial square root of zero.
All in all, you might well continue looking into it, but I can't guarantee that you will find a really nice, general result. But then again, if you cover many interesting classes, that's also something. ;)
